Showing the consistency of an equivalence relation over * Let $E$ be an equivalence relation on the set of all ordered pairs of non-negative integers ($N\times N$). It is defined as $$(a,b)E(x,y) \Longleftrightarrow a+y = b+x$$
Multiplication ($*$) is defined as $$(a,b)*(x,y) = (ax+by, ay+bx)$$
Without using substraction or division, how can I show that $E$ is consistent with $*$ ?
By consistent I mean that if $(a,b)E(a', b')$ and $(x,y)E(x',y')$ then $(a,b)*(x,y)E(a',b')*(x',y')$

Right now I have laid down all the information about the problem:
Let $(a,b)$, $(a',b')$, $(x,y)$ and $(x',y')$ $\in N \times N$. Suppose $(a,b)E(a',b')$ and $(x,y)E(x',y')$.
We have:
$$(a,b)E(a',b') \Longleftrightarrow a+b'= b+a'$$
$$(x,y)E(x',y') \Longleftrightarrow x+y'= y+x'$$
and I want to show that the following is true 
$$(a,b)*(x,y)E(a',b')*(x',y')$$
The expansion of the preceding statement gives this:
$$ax+by+b'x'+a'y'=^{?} bx+ay+a'x'+b'y'$$
I spent some time thinking about it but I can't find a way to do that.
Thanks for the help!
 A: What you have in front of you is actually a construction of the integers $\mathbb{Z}$ from the non-negative integers $N$. Define a surjection $\pi : N \times N \to \mathbb{Z}$ by $\pi(x,y) = x-y$. You should find that $(x,y) E (a,b)$ if and only if $\pi(x,y) = \pi(a,b)$. So, you can identity $E$-equivalence classes with integers!
Now, regarding consistency of $*$, observe that
$$\pi( (x,y)*(a,b)) = \pi(x,y) \pi(a,b)$$
where juxtaposition is ordinary multiplication of integers. Does this help you?

Added:
I did not see your request there be no reference to subtraction at first. So here is a more direct approach.

Claim 1: If $(a,b),(a',b'),(x,y) \in N \times N$ and  $(a,b) E (a',b')$, then $[(a,b)*(x,y)] E [(a',b') * (x,y)]$.
Proof: Since $(a,b)E(a',b')$, we have
$$ a+b' = b+a'.$$ Thus, we get
$$(a+b') x + (b+a')y = (b+a')x + (a+b')y$$
which expands out to
$$(ax+by) + (a'y+b'x) = (ay+bx)+(a'x+b'y)$$
which says exactly that
$$[(a,b)*(x,y)] E [(a',b') * (x,y)].$$

Similarly, you can prove

Claim 2: If $(a,b),(x,y),(x',y') \in N \times N$ and  $(x,y) E (x',y')$, then $[(a,b)*(x,y)] E [(a,b) * (x',y')]$.

Now, consistency of $*$ follows from transitivity of $E$, which I presume you have already checked? Transitivity of $*$ needs the cancellation law $x+a =y+a \Rightarrow x=y$ to have been established for $(N,+)$.
